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\begin{document}

\title{高等代数二}
\subtitle{02-最大公因式-因式分解 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月2日} }

\maketitle

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\begin{frame}{2.1. 作业：星期天晚上十点半之前在网络教学平台提交 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item   整理课堂笔记里的重点难点。补充没写完的计算或证明。
\item   习题(2.3)任选2题。抄写题目。
\item   习题(2.4)任选2题。抄写题目。
\end{enumerate}

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\begin{enumerate}

\item   公因式、最大公因式
\item   辗转相除法
\item   互素的两个多项式
\item   互素的充分必要条件
\item   互素的性质

\item   平凡因式
\item   可约多项式、不可约多项式
\item   不可约多项式的基本性质
\item   因式分解的存在性、唯一性
\item   典型分解式

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item   使用辗转相除法计算两个多项式的最大公因式。
\item   证明两个多项式互素的充分必要条件。
\item   理解不可约多项式的概念。
\item   证明因式分解的存在性与唯一性。
\item   写出多项式的典型分解式。

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. 最大公因式 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问： 设 $F$ 是一个数域，设 $f(x), g(x), h(x)\in F[x]$. 称 $h(x)$ 是 $f(x)$ 和 $g(x)$ 的一个公因式，指的是什么？}

\item  答：若 $h(x)\mid f(x)$ 且 $h(x)\mid g(x)$, 则称 $h(x)$ 是 $f(x)$ 和 $g(x)$ 的一个公因式。

\vspace{0.5cm}

\item  {\color{red}问： 什么是最大公因式？}

\item  答：称 $d$ 是 $f$ 和 $g$ 的最大公因式，是指：
\begin{enumerate}
\item  $d\mid f$ 且 $d\mid g$.
\item  若 $h\mid f$ 且 $h\mid g$, 则 $h\mid d$. 
\end{enumerate}

\vspace{0.5cm}

\item  {\color{red}记号：最高次项系数是1的那个最大公因式记为 $$(f(x),g(x)).$$}


\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：求 $(x^2-1)$ 和 $(x^3-1)$ 的所有公因式。 } 

\item  解答：对任意 $c\in F, c\neq 0$, $c$ 与 $c(x-1)$ 都是 $(x^2-1)$ 和 $(x^3-1)$ 的公因式。而且 $c(x-1)$ 是 $(x^2-1)$ 和 $(x^3-1)$ 的最大公因式。


\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.3.1. 数域 $F$ 上的两个多项式 $f(x)$ 与 $g(x)$ 一定存在最大公因式，而且任意两个最大公因式之间只相差一个零次因式。}

\item  证明：
\begin{enumerate}
\item  存在性。使用辗转相除法求得最大公因式。
\item  唯一性。研究辗转相除法得到的一系列等式。


\end{enumerate}

\end{itemize}

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\begin{frame}{2.6. 辗转相除法 }

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\begin{itemize}

\item  {\color{red}问：设有如下辗转相除法的结果。求 $f(x)$ 与 $g(x)$ 的最大公因式。}
\begin{eqnarray*}
f(x) &=& g(x)q_1(x) + r_1(x),\quad \deg(r_1)<\deg(g), \\
g(x) &=& r_1(x)q_2(x) + r_2(x), \quad \deg(r_2)<\deg(r_1),\\
r_1(x) &=& r_2(x)q_3(x) + r_3(x), \quad \deg(r_3)<\deg(r_2),\\
r_2(x) &=& r_3(x)q_4(x).
\end{eqnarray*}

\item  答：$f(x)$ 与 $g(x)$ 的一个最大公因式是 $r_3(x)$. 其它最大公因式都有 $cr_3(x)\, (c\in F,c\neq 0)$ 的形式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：求多项式 $(x^2-1)$ 与 $(x^3-1)$ 的最大公因式。}

\item  解答：这两个多项式的最大公因式可以写成 $$c(x-1)\quad (c\in F,c\neq 0)$$ 的形式，而首项系数为1的最大公因式只有1个，即 $$(x^2-1,x^3-1) = x-1. $$

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：在有理系数范围内，求下述两个多项式的最大公因式：
\begin{eqnarray*}
 f(x) &=& x^4-2x^3-4x^2+4x-3, \\
 g(x) &=& 2x^3-5x^2-4x+3.
\end{eqnarray*}
}

\item  答：辗转相除法。为避免分式运算，先将被除式乘以适当的整数。
\begin{eqnarray*}
 4 f(x) &=& q_1(x)g(x) + r_1(x), \\ 
 9 g(x) &=& q_2(x)r_1(x) + r_2(x), \\ 
 56 r_1(x) &=& q_3(x)r_2(x). 
\end{eqnarray*}
其中 $q_1(x)=2x+1$, $r_1(x)=-3x^2+14x-15$, $q_2(x)=-6x-13$, $r_2(x)=56x-168$, $q_3(x)=-3x+5$. 
可得最大公因式为 $r_2(x)$ 或 $x-3$. 


\end{itemize}

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\begin{itemize}

\item  可以用R语言里的程序包 pracma 计算多项式的加减乘除运算。

\begin{lstlisting}[language=R]
library(pracma)
f=c(1,-2,-4,4,-3)
g=c(2,-5,-4,3)
res1=polydiv(4*f,g)
q1=res1$d
r1=res1$r
res2=polydiv(9*g,r1)
q2=res2$d
r2=res2$r
res3=polydiv(56*r1,r2)
q3=res3$d
r3=res3$r
\end{lstlisting}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.3.2. 设 $d(x)$ 是数域 $F$ 上的多项式 $f(x)$ 与 $g(x)$ 的最大公因式，则存在数域 $F$ 上的多项式 $u(x),v(x)$ 使得 $$f(x)u(x) + g(x)v(x)=d(x).$$ }

\item  证明：写出辗转相除法的一系列等式。从后往前写，可以将 $d(x)$ 写成 $f(x)$ 与 $g(x)$ 的组合的形式。

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：在有理系数范围内，求下述两个多项式的最大公因式 $d(x)$, 
\begin{eqnarray*}
 f(x) &=& x^4-2x^3-4x^2+4x-3, \\
 g(x) &=& 2x^3-5x^2-4x+3.
\end{eqnarray*}
并求 $u(x),v(x)$ 使得 $f(x)u(x) + g(x)v(x)=d(x)$. 
}

\vspace{0.3cm}

\item  例子的解答：写出辗转相除法的一系列等式，
\begin{eqnarray*}
 4 f(x) &=& q_1(x)g(x) + r_1(x), \\ 
 9 g(x) &=& q_2(x)r_1(x) + r_2(x), \\ 
 56 r_1(x) &=& q_3(x)r_2(x). 
\end{eqnarray*}
于是可知 $d(x) = r_2(x)$ 是 $f(x)$ 与 $g(x)$ 的最大公因式。

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  从第二个等式往回代，可得 
\begin{eqnarray*}
r_2 &=& 9g-q_2r_1 \\ 
&=& 9g - q_2(4f-q_1g) \\ 
&=& -4q_2f + (9+q_2q_1)g.
\end{eqnarray*}

\item  因此所求 $u, v$ 可以取为 $u=-4q_2$, $v=9+q_2q_1$.

\item  注：如果要求 $d(x)$ 是首项系数为1的那个最大公因式，即 
$$d(x) = r_2(x)/56,$$
则取 $u=-4q_2/56, v=(9+q_2q_1)/56.$ 

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问： 什么时候称两个多项式为互素？}

\item  答：当这两个多项式的最大公因式是零次因式的时候。即
$$(f(x),g(x))=1.$$

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：证明 $\mathbb{Q}[x]$ 中的两个多项式 $4(x^3+1)$ 与 $6(x^2+1)$ 互素。}

\item  证明：这两个多项式的最大公因式是非零常数，即零次因式。因此按定义可得它们是互素的。

%\item  注：
%\begin{enumerate}
%\item  在整数环 $\mathbb{Z}$ 中，我们有 $(4,6)=2$. 它们不是互素的。
%\item  在多项式环 $\mathbb{R}[x]$ 中，我们有 $(4(x^3+1),6(x^2+1))=1$. 它们是互素的。
%\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.3.3. 两个多项式 $f(x),g(x)$ 互素当且仅当存在两个多项式 $u(x),v(x)$ 使得
$$f(x)u(x)+g(x)v(x) =1. $$}

\item  证明：按照互素的定义，和上一个定理。

\end{itemize}

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\begin{itemize}
\item  {\color{red}问题：写出互素的一些性质，并加以证明。}

\vspace{0.3cm}

\item  解答：基本性质有下述几个。
\begin{enumerate}
\item  若 $f$ 和 $g$ 都与 $h$ 互素，则 $fg$ 也与 $h$ 互素。
\item  若 $h$ 整除 $fg$, 且 $h$ 与 $f$ 互素，则 $h$ 整除 $g$. 
\item  若 $g\mid f$, $h\mid f$ 且 $g$ 与 $h$ 互素，则 $gh\mid f$. 
\end{enumerate}

证明从整除与互素的定义出发。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是一个多项式的平凡因式？}

\item  答：多项式 $f(x)$ 的平凡因式是指 $c$ 和 $cf(x)$, 其中 $c\in F, c\neq 0$. 


\item  注：下述三种陈述，说的是同一个意思：
\begin{enumerate}
\item  $f(x)=c$, 其中 $c\in F, c\neq 0$.
\item  $f(x)$ 是零次多项式。
\item  $f(x)$ 是非零常数。 
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是可约的多项式？}
\item  答：设 $f(x)$ 是一个次数大于等于1的多项式。
%有两种等价的定义：
%\begin{enumerate}
%\item  如果 $f(x)$ 除了平凡因式，还有其它因式，则称为可约的。
%\item  
如果存在两个次数都比 $f(x)$ 的次数小的多项式 $g(x)$ 和 $h(x)$ 使得 $$f(x)=g(x)h(x)$$, 那么称 $f(x)$ 为可约的。
%\end{enumerate}

\vspace{0.5cm}

\item  {\color{red}问：什么是不可约的多项式？}
\item  答：设 $f(x)$ 是一个次数大于等于1的多项式。
%有两种等价的定义：
%\begin{enumerate}
%\item   一个多项式如果不是可约的，就称为是不可约的。
%\item   如果 $f(x)$ 写成两个多项式的乘积，那么这两个都是平凡因式。 
%\end{enumerate}
如果 $f(x)$ 不能写成两个次数较低的多项式的乘积，那么称 $f(x)$ 为不可约的。

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：解释下述结论。}
\begin{enumerate}
\item   {\color{red}在任意数域 $F$ 上, 一次多项式 $a_1x+a_0\, (a_1\neq 0)$ 都是不可约的。}
\item   {\color{red}在有理数域 $\mathbb{Q}$ 上，$x^2-2$ 是不可约的。}
\item   {\color{red}在实数域 $\mathbb{R}$ 上，$x^2-2$ 是可约的。}
\end{enumerate}

\item  解答：验证可约多项式与不可约多项式的定义。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：证明不可约多项式的一些基本性质：
\begin{enumerate}
\item   {\color{red}若 $p(x)\in F[x]$ 不可约，则对任意 $c\in F-\{0\}$, $cp(x)$ 也不可约。}
\item   {\color{red}若  $p(x)\in F[x]$ 不可约，则对任意 $f(x)\in F[x]$, 要么 $p\mid f$, 要么 $p$ 与 $f$ 互素。}
\item   {\color{red}若  $p(x)\in F[x]$ 不可约，设 $f(x),g(x)\in F[x]$, 若 $p\mid fg$, 则 $p\mid f$ 或 $p\mid g$. }
\end{enumerate}
}

\item  解答：从不可约多项式的定义出发。

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}定理2.4.1. 设 $F$ 是一个数域。则 $F[x]$ 中的任意 $n$ 次多项式 $(n\ge 1)$ 都能写成 $F[x]$ 中的一些不可约多项式的乘积。}

\item  证明：%（使用数学归纳法的写法）
\begin{enumerate}
\item   当 $f(x)$ 的次数是1，它一定是不可约的。
\item   归纳假设：当 $f(x)$ 的次数小于等于 $n$ 时，$f(x)$ 都能写成一些不可约多项式的乘积。
\item   现设 $f(x)$ 的次数等于 $n+1$.
\item   若 $f(x)$ 不可约，则已得证。
\item   若 $f(x)$ 可约，则 $f(x)=f_1(x)f_2(x)$, 其中 $f_1(x)$ 和 $f_2(x)$ 的次数都比 $f(x)$ 的次数小。对这两个因式使用归纳假设，这两个因式都可以写成一些不可约多项式的乘积。于是 $f(x)$ 也可以写成一些不可约多项式的乘积。
\end{enumerate}

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：}

\begin{enumerate}
\item  {\color{red}将下述整数写成一些素数的乘积：}
$${\color{red}123456.}$$ 
\item  {\color{red}将下述多项式写成一些不可约多项式的乘积：}
$$ {\color{red} f(x) = 6x^6+5x^5+4x^4+3x^3+2x^2+x+1.} $$ 
\end{enumerate}

\item  答：由因式分解定理，所求因式分解是存在的。但是很难找到它们。


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}定理2.4.2. 设 $F$ 是一个数域。设 $f(x)\in F[x]$ 是任意一个 $n$ 次多项式，$n\ge 1$. 若有两个不可约因式分解
\begin{eqnarray*}
f(x) = p_1(x)p_2(x)\cdots p_r(x) = q_1(x)q_2(x)\cdots q_s(x).
\end{eqnarray*}
则 $r=s$, 且存在指标集 $\{1,2,\cdots,r\}$ 的一个置换 $\sigma$, 以及一些非零系数 $c_1,c_2,\cdots,c_r$, 使得
$$q_i(x) = c_ip_{\sigma(i)}(x),\quad i=1,2,\cdots,r.$$
}


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\begin{itemize}

\item  {\color{red}问题：写出 $x^2-1$ 的两种不同的因式分解。}

\item  解答：不同的因式分解方式，仅相差前后顺序和非零常数。%这里 $\sigma(1)=2$, $\sigma(2)=1$, $c_1=2$, $c_2=1/2$.  
\begin{eqnarray*}
x^2-1 = (x+1)(x-1) = (2x-2)(\frac{1}{2}x+\frac{1}{2}).
\end{eqnarray*}

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\begin{itemize}

\item  {\color{red}证明因式分解的唯一性的思路：}

\begin{enumerate}
\item   先考虑一个最简单的情形。设有
\begin{eqnarray*}
f(x) = p_1(x)p_2(x) = q_1(x)q_2(x),
\end{eqnarray*}
其中 $p_1,p_2,q_1,q_2$ 都是不可约的多项式。怎么证明 $q_1(x)=c_1p_1(x)$ 或 $q_1(x)=c_1p_2(x)$ 呢？

\item   由 $p_1\mid q_1q_2$ 以及 $p_1$ 不可约，得出 $p_1\mid q_1$ 或 $p_1\mid q_2$.

\item  若 $p\mid q$ 且 $q\mid p$, 则 $p$ 与 $q$ 只相差一个零次因式。

\end{enumerate}

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\begin{itemize}

\item  {\color{red}问：什么是典型分解式？}

\item  答：把不可约因式都写成首项系数为1，把相同的不可约因式写成幂次，
$$f(x)=ap_1(x)^{k_1}p_2(x)^{k_2}\cdots p_t(x)^{k_t},$$
这里 $p_1,p_2,\cdots,p_t$ 是互不相同的首项系数为1的不可约多项式。
\vspace{0.5cm}

\item  注：如果有了典型分解式，就很容易写出最大公因式了。只要找相同的不可约因式，然后观察幂次，选取较小的幂次就可以了。

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\begin{itemize}


\item   {\color{red}问题：计算下述两个多项式的最大公因式，
\begin{eqnarray*}
f(x) &=& x^4+3x^3-x^2-4x-3, \\
g(x) &=& 3x^3+10x^2+2x-3.
%\label{eq-2-3-1}
\end{eqnarray*}
}

\item   解答：使用辗转相除法，可得 $(f(x),g(x)) = x+3$.


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\begin{itemize}


\item   {\color{red}问题：设 $f(x)$ 与 $g(x)$ 是 $F[x]$ 的多项式，而 $a,b,c,d$ 是 $F$ 中的数，并且 $ad-bc\neq 0$. 证明 
\[ (af(x)+bg(x), cf(x)+dg(x)) = (f(x),g(x)). \]
}

\item   解答：记 $d=(f,g)$. 按定义验证。
\begin{enumerate}
\item  证明 $d$ 是 $af+bg$ 和 $cf+dg$ 的公因式。
\item  设 $h$ 是 $af+bg$ 和 $cf+dg$ 的公因式。证明 $h\mid d$. 
\end{enumerate}



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\begin{itemize}

\item  {\color{red}问题：考虑下述两个有理系数多项式
\begin{eqnarray*}
f(x) &=& x^4+2x^3-x^2-4x-2, \\
g(x) &=& x^4+x^3-x^2-2x-2.
%\label{eq-2-3-5}
\end{eqnarray*}
求有理系数多项式 $u(x)$ 与 $v(x)$, 使得下述等式成立，
\[ u(x)f(x) + v(x)g(x) = (f(x),g(x)). \]
}

\item   解答：先写出辗转相除法的每个等式。$(f,g)=x^2-2$. 


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\begin{itemize}


\item  {\color{red}问题：设 $(f,g)=1$. 证明：$(f,f+g)=1$, $(fg, f+g)=1$. }

\item   解答：使用互素的充分必要条件。


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\begin{itemize}

\item  {\color{red}问题：证明 $(f,g)^2 = (f\,^2,g^2)$. }

\item   解答：考虑最大公因式的定义。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：确定 $k$, 使得下述两个多项式的最大公因式是一次的：
\begin{eqnarray*}
f(x) &=& x^2 + (k+6)x +4k+2, \\
g(x) &=& x^2 + (k+2)x +2k.
%\label{eq-2-3-5}
\end{eqnarray*}
}

\item   解答：使用辗转相除法。$k=1$ 或 $k=3$. 


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\begin{itemize}

\item   {\color{red}问题：在有理数域上把下述多项式分解为一些不可约多项式的乘积，}
\begin{enumerate}
\item  {\color{red}$f(x) = 3x^2+1$. }
\item  {\color{red}$f(x) = x^3-2x^2-2x+1$. }
\end{enumerate}


\item   解答：
\begin{enumerate}
\item  不可约。
\item  观察到 $x=-1$ 时 $f(x)=0$. 
\end{enumerate}

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\begin{itemize}

\item   {\color{red}问题：分别在复数域、实数域和有理数域上分解多项式 $x^4+1$ 为不可约多项式。}

\item   解答：先在复数域上分解因式。考虑单位根。


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\begin{itemize}

\item   {\color{red}问题：证明 $g(x)^2\mid f(x)^2$ 当且仅当 $g(x)\mid f(x)$. }

\item   解答：
\begin{enumerate}
\item  设 $g(x)\mid f(x)$. 考虑整除的定义。
\item  设 $g(x)\nmid f(x)$. 考虑 $f(x)$ 除以 $g(x)$ 得到的商和余式。
\end{enumerate}

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\begin{itemize}

\item   {\color{red}问题：}
\begin{enumerate}
\item  {\color{red}求 $f(x)=x^5-x^4-2x^3+2x^2+x-1$ 在 $\mathbb{Q}[x]$ 中的典型分解式。}
\item  {\color{red}求 $f(x)=2x^5-10x^4+16x^3-16x^2+14x-6$ 在 $\mathbb{R}[x]$ 中的典型分解式。}
\end{enumerate}

\item   解答：
\begin{enumerate}
\item  测试 $x=\pm 1$ 时 $f(x)$ 是否等于零，即猜测 $f(x)$ 是否有因式 $x\pm 1$. 
\item  先提出公因子2，然后测试 $x=\pm 1, \pm 2, \pm 3$ 时 $f(x)$ 是否等于零。 
\end{enumerate}

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\begin{itemize}

\item   {\color{red}问题：设 $f(x)$ 是数域 $F$ 上的一个次数大于零的多项式。证明下述两个条件相互等价。}
\begin{enumerate}
\item[(i)]   {\color{red}存在 $F[x]$ 中的一个不可约多项式 $p(x)$ 以及一个正整数 $k$ 使得} 
$$ {\color{red} f(x)=p(x)^k. } $$  
\item[(ii)]  {\color{red}对任意 $g(x)\in F[x]$, 或者 $(f(x),g(x))=1$, 或者存在一个正整数 $m$ 使得} 
$$ {\color{red} f(x)\mid g(x)^m. } $$  
\end{enumerate}

\item   解答：
\begin{enumerate}
\item  (i) $\Rightarrow$ (ii): 考虑 $g(x)$ 的典型分解式。
\item  (ii) $\Rightarrow$ (i): 考虑 $f(x)$ 的典型分解式。
\end{enumerate}




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